Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,791 comments
1,470 users with positive rep
820 active unimported users
More ...

  Dimensional reduction of Yang-Mills to m(atrix) theory

+ 4 like - 0 dislike
1065 views

The Yang-Mills action are usually given by $$S= \int\text{d}^{10}\sigma\,\text{Tr}\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\theta^{T}\gamma^{\mu}D_{\mu}\theta\right)$$

with the field strength defined as $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}-ig\left[A_{\mu},A_{\nu}\right]$ , $A_{\mu}$ being a U(N) Hermitian gauge field in the adjoint representation, $\theta$ being a $16\times1$ Majorana-Weyl spinor of $SO(9)$ in the adjoint representation and $\mu=0,\dots,9$ . The covariant derivative is given by $D_{\mu}\theta=\partial_{t}\theta-ig\left[A_{\mu},\theta\right]$. We are using a metric with mostly positive signs.
 

We re-scale the fields by $A_{\mu}\to\frac{i}{g}A_{\mu}$ and let $g^{2}\to\lambda$ which gives us $$S=\int\text{d}^{10}\sigma\,\text{Tr}\left(\frac{1}{4\lambda}F_{\mu\nu}F^{\mu\nu}-\theta^{T}\gamma^{\mu}D_{\mu}\theta\right)$$ with the field strength defined as $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}+\left[A_{\mu},A_{\nu}\right]$ and the covariant derivative $D_{\mu}\theta=\partial_{t}\theta+\left[A_{\mu},\theta\right]$.

Now we perform a dimensional reduction from $9+1$ to $0+1$ , so that all the fields only depend on time, thous all spatial derivatives vanish i.e. $\partial_{a}(\text{Anything})=0$ . The $10$ -dimensional vector field decomposes into $9$ scalar fields $A_{a}$ which we rename $X^{a}$ and one gauge field $A_{0}$ which we rename $A$ . This gives (note that $\gamma^{t}=\mathbb{I}$ and that $\gamma^{a}=\gamma_{a}$. $$F_{0a}= \partial_{t}X^{a}+\left[A,X^{a}\right],\quad F_{ab}=+\left[X^{a},X^{b}\right] \gamma^{t}D_{t}\theta= \partial_{t}\theta+\left[A,\theta\right],\quad\gamma^{a}D_{a}\theta=\gamma_{a}\left[X^{a},\theta\right]$$

The action for this theory is then $$S=\int\text{d}t\,\text{Tr}\left(\frac{1}{2\lambda}\bigg\{-\left(D_{t}X^{a}\right)^{2}+\frac{1}{2}\left[X^{a},X^{b}\right]^{2}\bigg\}-\theta^{T}D_{t}\theta-\theta^{T}\gamma_{a}\left[X^{a},\theta\right]\right)$$ with the covariant derivative defined as $D_{t}X^{a}=\partial_{t}X^{a}+\left[A,X^{a}\right]$ and $D_{t}\theta=\partial_{t}\theta+\left[A,\theta\right]$

Now to the question. I need the potential energy $V=+\frac{1}{2}\left[X^{a},X^{b}\right]^{2}$ to be negative, not positive.

Taylor has a discussion on this in his paper "Lectures on D-branes, Gauge Theory and M(atrices)" (http://arxiv.org/abs/hep-th/9801182) on page 10, where he writes:
"Because the metric we are using has a mostly positive signature, the kinetic terms have a single raised 0 index corresponding to a change of sign, so the kinetic terms indeed have the correct sign. The commutator term $\left[X^{a},X^{b}\right]^{2}$ which acts as a potential term is actually negative definite. This follows from the fact that $\left[X^{a},X^{b}\right]^{\dagger}=\left[X^{b},X^{a}\right]=-\left[X^{a},X^{b}\right]$. Thus, as expected, kinetic terms in the action are positive while potential terms are negative."

But I don't understand where the Hermitian conjugate $^\dagger$ comes from, to me this term is just: $$\left[X^{a},X^{b}\right]^{2}=\left[X^{a},X^{b}\right]\left[X_{a},X_{b}\right]$$

Note that Taylor uses a little different conventions when he re-scales, instead of $A_{\mu}\to\frac{i}{g}A_{\mu}$ he uses $A_{\mu}\to\frac{1}{g}A_{\mu}$ and $\theta \to\frac{1}{g}\theta$. But this should not cause any troubles I think.


This post imported from StackExchange Physics at 2014-03-07 16:47 (UCT), posted by SE-user Natanael

asked Nov 10, 2013 in Theoretical Physics by Natanael (75 points) [ revision history ]
retagged May 21, 2014 by dimension10

1 Answer

+ 2 like - 0 dislike

In the first action the $A_{\mu}$ are Hermitian.
In the second action the $A_{\mu}$ are anti-Hermitian since we let $A_{\mu}\to\frac{i}{g}A_{\mu}$. The commutator of anti-Hermitian matrices are also anti-Hermitian.
If we have $\text{Tr}(M^{2})$ , with $M$ being anti-Hermitian, then we can write it as $\text{Tr}(M^{2})=-\text{Tr}((iM)^{2})$ , with $iM$ being Hermitian. Since the eigenvalues of an Hermitian matrix is real and we take the trace of the square it, it follows that $\text{Tr}(M^{2})\leq0$ . Changing the anti-Hermitian matrices to Hermitian matrices changes the sign.

This post imported from StackExchange Physics at 2014-03-07 16:47 (UCT), posted by SE-user Natanael
answered Nov 22, 2013 by Natanael (75 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...